Calculus I · Adding Up Forever

Principle 1 — Comparison: trap it between sums you know

The first principle is exactly what you’d do by instinct: if your series is term-by-term smaller than one you know converges, it must converge too; if it’s larger than one you know diverges, it must diverge.

Comparison. Suppose $0 \le a_n \le b_n$ for all $n$. If $\displaystyle\sum b_n$ converges, then $\displaystyle\sum a_n$ converges. If $\displaystyle\sum a_n$ diverges, then $\displaystyle\sum b_n$ diverges.

The geometric series is usually the thing you compare against. For instance, $\displaystyle\sum \frac{1}{2^n + 1}$ has every term smaller than $\tfrac{1}{2^n}$, and $\sum \tfrac{1}{2^n}$ is a convergent geometric series — so by comparison, $\sum \tfrac{1}{2^n+1}$ converges. The whole skill is recognizing what to compare to: bound your complicated series above by a simple convergent one (to prove convergence), or below by a simple divergent one (to prove divergence). (Figure 12.2: two stacks of bars, the $a_n$ bars sitting entirely under the $b_n$ bars, with the $b_n$ total marked finite — so the $a_n$ total is trapped finite beneath it.)

Principle 2 — The integral test: weigh the sum against an area

The second principle is the one that earns this chapter its place at the end of a calculus book, because it connects a sum to an integral — and it does so with the very rectangle-and-curve picture you built in Chapters 10 and 11.

Take a series $\sum f(n)$ whose terms come from a function $f$ that is positive, decreasing, and continuous. Draw the terms as rectangles of width $1$ and height $f(n)$, and lay them against the graph of $f$. Because $f$ is decreasing, each rectangle is trapped between the area just under and just over the curve — and summing across all of them traps the whole sum between two copies of the same integral:

$$\sum_{n=2}^\infty f(n) \;\le\; \int_1^\infty f(x)\,dx \;\le\; \sum_{n=1}^\infty f(n).$$

The sum and the integral differ by at most the single term $f(1)$, so they are finite together or infinite together. That is the test:

The integral test. For positive, decreasing, continuous $f$, the series $\displaystyle\sum_{n=1}^\infty f(n)$ and the integral $\displaystyle\int_1^\infty f(x)\,dx$ either both converge or both diverge.

Look at what just happened: the integral you learned to compute with the Fundamental Theorem now decides infinite sums. The rectangles that defined the integral in Chapter 10 reappear here as the terms of a series, and the area they approximate settles whether the sum is finite. Adding up infinitely many numbers and measuring the area under a curve are, once again, the same question. (Figure 12.3: the curve $y = f(x)$ with the series’ rectangles drawn against it — one set tucked under the curve, one rising above — visibly squeezing the sum to the area.)

The p-series, settled at last — and a debt from Chapter 11 repaid

Apply the integral test to the family $\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}$. Convergence hinges on $\displaystyle\int_1^\infty \frac{1}{x^p}\,dx$ — which we computed in Chapter 11: it is finite exactly when $p > 1$. Therefore:

$\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}$ converges if and only if $p > 1$.

This resolves the cliffhanger Chapter 11 left you with. There, $\int_1^\infty \tfrac1x\,dx$ was infinite while $\int_1^\infty \tfrac{1}{x^2}\,dx$ was finite — and you were asked to predict which of $\sum\tfrac1n$ and $\sum\tfrac1{n^2}$ converges. Now you know, and you know why they match: the harmonic series ($p=1$) diverges because its integral does; $\sum \tfrac{1}{n^2}$ ($p=2$) converges because its integral does. (That sum, remarkably, equals $\tfrac{\pi^2}{6}$ — a fact whose proof waits for the Fourier methods of Book 3.) The sum and the area live or die together, exactly as Chapter 11 promised.

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